Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(D, app(app(-, x), y)) → APP(D, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

APP(D, app(app(-, x), y)) → APP(D, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP(D, app(app(-, x), y)) → APP(D, y)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, y))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 19 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(*, x), y)) → APP(D, y)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

D1(+(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
D1(-(x, y)) → D1(y)
D1(+(x, y)) → D1(y)
D1(-(x, y)) → D1(x)

R is empty.
The set Q consists of the following terms:

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
map(x0, nil)
map(x0, cons(x1, x2))
filter(x0, nil)
filter(x0, cons(x1, x2))
filter2(true, x0, x1, x2)
filter2(false, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.


The following pairs can be oriented strictly and are deleted.


APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(*, x), y)) → APP(D, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D1(x1)  =  x1
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x1, x2)
-(x1, x2)  =  -(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The remaining pairs can at least be oriented weakly.

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.